直线 $3x - 5y = 10$ 穿过点 A $(5, 1)$ 且与 y 轴交于点 B。

1. Rewrite the line equation in gradient-intercept form and identify the gradient.

$3x-5y=10$
$-5y=10-3x$
$y=\frac{3}{5}x-2$
Comparing with $y=mx+c$, the gradient is $\dfrac{3}{5}$.

1. Find the coordinates of point B on the y-axis.

Since B is on the y-axis, $x=0$.

Substitute into $3x-5y=10$:
$3(0)-5y=10$
$-5y=10$
$y=-2$
So $B=(0,-2)$.

2. Use the distance formula to find the length of AB.

A is $(5,1)$ and B is $(0,-2)$.

$AB=\sqrt{(5-0)^2+(1-(-2))^2}$
$=\sqrt{25+9}$
$=\sqrt{34}$

1. Find the y-coordinate of B and express the length BC in terms of k.

Since B is on the y-axis, set $x=0$ in $3x-5y=10$.

$-5y=10$
$y=-2$

So $B=(0,-2)$.

Given $C=(0,k)$, the vertical length
$BC=|k-(-2)|=|k+2|.$

2. Use the area of triangle ABC to solve for k.

Take $BC$ as the base. Since $BC$ lies on the y-axis, the perpendicular height from $A(5,1)$ to the y-axis is $5$ units.

$\text{Area}=\frac12\times BC\times \text{height}$

$10=\frac12\times |k+2|\times 5$

$|k+2|=4$

So
$k+2=4 \quad \text{or} \quad k+2=-4$

$k=2 \quad \text{or} \quad k=-6.$